# 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

## Problem

Suppose $A, R, S$, and $T$ all denote distinct digits from $1$ to $9$. If $\sqrt{STARS} = SAT$, what are $A, R, S$, and $T$?

## Solution

$A = 3, T = 9, S = 1$ and $R = 2$

Another way of stating the question is that $SAT$ $\cdot$ $SAT$ = $STARS$. This means that the ending digit of $T$ $\cdot$ $T$ or $S$ has to be the ending digit of a square number, or {$0, 1, 4, 5, 6, 9$}. Keep in mind that $STARS$ is a 5-digit number. We know that $S$ cannot be 0 because $SAT$ would become a 2-digit number. It cannot be 4 or over because $STARS$ would have 6 digits instead of 5. This only leaves $S$ = 1. Because the last digit of $STARS$ is $1$, we can conclude that $T$ must be $9$ or $1$. However, since all the digits are unique, $T$ = $9$. Now the expression becomes:

$1A9$ $\cdot$ $1A9$ = $19AR1$

Because the first two digits of $STARS$ are $1$ and $9$, $A$ cannot take on too many values because if it becomes $5$, $S$ will become $2$. It cannot be $4$ either because even then, $STARS$ would be too small to satisfy (If $SAT=140$, it would still have too big a square for $STARS$ to be able to satisfy, $A$ being $4$). Now, if $A$ is tested from the set {$0,1,2,3$}:

$A=0$:

$109$ $\cdot$ $109$ = $11881$ (Does not work)

$A=1$:

$119$ $\cdot$ $119$ = $14161$ (Does not work)

$A=2$:

$129$ $\cdot$ $129$ = $16641$ (Does not work)

$A=3$:

$139$ $\cdot$ $139$ = $19321$

This works as it fits the template: $\boxed{\textbf{(S=1, T=9, A=3, and R=2)}}$

## See also

 2017 UNM-PNM Contest II (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNM-PNM Problems and Solutions